Difference between revisions of "009B Sample Midterm 1, Problem 2"

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<span class="exam">Find the average value of the function on the given interval.
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<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
  
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::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>
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<span class="exam">and finds that it can be approximated by the function
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::::::<math>s(t)=t(25-5t)+18</math>
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<span class="exam">where <math>t</math> is the time (in years) since the stock was purchased.
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<span class="exam">Find the average price of the stock over the first five years.
  
  
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|Using the formula given in Foundations, we have:
 
|Using the formula given in Foundations, we have:
 
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| &nbsp; &nbsp;<math style="vertical-align: 0px">f_{\text{avg}}=\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx=\int_0^2 x^3(1+x^2)^4~dx.</math>  
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| &nbsp; &nbsp;<math style="vertical-align: 0px">f_{\text{avg}}=</math>  
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we use <math>u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^2</math>. Then, <math style="vertical-align: 0px">du=2x dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=xdx</math>. Also, <math style="vertical-align: 0px">x^2=u-1</math>.
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|We need to change the bounds on the integral. We have <math style="vertical-align: -4px">u_1=1+0^2=1</math> and <math style="vertical-align: -3px">u_2=1+2^2=5</math>.
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|So, the integral becomes <math style="vertical-align: -19px">f_{\text{avg}}=\int_0^2 x\cdot x^2 (1+x^2)^4~dx=\frac{1}{2}\int_1^5(u-1)u^4~du=\frac{1}{2}\int_1^5(u^5-u^4)~du</math>.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
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|We integrate to get
 
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| &nbsp; &nbsp; <math>f_{\text{avg}}=\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5=\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.</math>
 
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!Step 4: &nbsp;
 
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|We evaluate to get
 
 
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| &nbsp; &nbsp; <math style="vertical-align: -20px">f_{\text{avg}}=5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)=3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}=\frac{59376}{60}=\frac{4948}{5}</math>.
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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| &nbsp; &nbsp; <math>\frac{4948}{5}</math>
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| &nbsp; &nbsp; <math></math>
 
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:14, 5 February 2017

Otis Taylor plots the price per share of a stock that he owns as a function of time

and finds that it can be approximated by the function

where is the time (in years) since the stock was purchased.

Find the average price of the stock over the first five years.


Foundations:  
The average value of a function on an interval is given by .

Solution:

Step 1:  
Using the formula given in Foundations, we have:
   
Step 2:  
Step 3:  
Final Answer:  
   

Return to Sample Exam

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